# $n$-th homology of open subset of $\mathbb{R}^n$

It holds that for an open subset $$U\in\mathbb{R}^n$$ the $$n$$-th singular homology group $$H_n(U)$$ is trivial. A proof can be found on page $$147$$ of Homology Theory: an Introduction to Algebraic Topology by James W. Vick.

I am looking for other references the prove this, hopefully in a simpler or more transparent way. Vick uses some notion of barycentric subdivision. I am not sure if this can be avoided but it would be interesting to see a different approach.

Sketch of the proof: For a cycle $$z$$ representing some element in $$H_n(U)$$ its "image" is supported by a compact (bounded) set $$X$$ so we can take an $$n$$-simplex in $$\mathbb{R}^n$$ that contains $$X$$. Using barycentric subdivision we can cut up this simplex into pieces of diameter smaller than $$\inf \{\lvert\lvert x-y\rvert\rvert, x\in X,y\in\mathbb{R}^n\setminus U\}$$

He considers the barycentric subdivision $$B$$ as a finite CW complex under the simplicial composition (not sure what this means) and takes a subcomplex $$K$$ consisting of all faces of simplices intersecting $$X$$. An argument on the long exact sequence of $$(B,K)$$ then shows that $$H_n(K)=0$$. This completes the proof since $$X\subset K\subset U$$

• Can you sketch the proof of Vick to avoid people giving you the same proof?
– Pedro
May 20, 2021 at 16:16
• Good idea! I added it. May 20, 2021 at 16:50

An open subset $$U$$ of $$\mathbb{R}^n$$ is a manifold. Non-compact Poincare Duality identifies $$H_*(U)$$ with $$\bar H^{n-*}(U^+)$$ where $$U^+$$ denotes the one point compactification. Letting $$*=n$$ we get that $$H_n(U)$$ is isomorphic with $$\bar{H}^0(U^+ )$$. Since $$U$$ is noncompact $$U^+$$ is path connected, so $$\bar{H}^0 (U^+ )=0$$.

So you might just look for a reference for noncompact Poincare duality. Hatcher does a pretty good job.

• Thanks for the suggestion. Vick shows Poincare Duality for compact manifolds and uses the statement as a lemma. Does Hatcher avoid this in his buildup for the non-compact case? May 20, 2021 at 16:56
• @JarneRenders Check Proposition 3.29 in Hatcher. It doesn't use simplicial complex structures, but Vick's POV has some merit. May 20, 2021 at 17:05

Here is a sketch of a more intuitive geometric approach, though the technical details get fairly involved. The basic idea is to triangulate $$U$$ and use simplicial homology, so that a nontrivial class in $$H_n(U)$$ would be forced to be a linear combination of fundamental classes of connected components of $$U$$ (where the fundamental class of a component is just the sum of all the $$n$$-simplices in that component), which is not actually possible since $$U$$ is not compact so this would be an infinite sum of simplices.

As a first step, you prove there exists a triangulation of $$U$$ by linear simplices in $$\mathbb{R}^n$$. There are various ways to prove this; for instance, you can take a sequence of increasingly fine cubical meshes on $$\mathbb{R}^n$$ and use this to write $$U$$ as a nice union of cubes, and then subdivide those cubes into simplices to get a triangulation.

Now since $$U$$ is an $$n$$-manifold, this triangulation has the property that each $$(n-1)$$-simplex is a face of exactly two $$n$$-simplices (if an $$(n-1)$$-simplex was a face of a different number of $$n$$-simplices, then $$U$$ would not be locally homeomorphic to $$\mathbb{R}^n$$ at a point in the interior of that $$(n-1)$$-simplex). This means that if $$\alpha$$ is a simplicial $$n$$-cycle with respect to this triangulation that has a nonzero coefficient on a certain $$n$$-simplex $$\sigma$$, then $$\alpha$$ is forced to also have a nonzero coefficient on each $$n$$-simplex that shares a face with $$\sigma$$, since that is the only way to cancel out each boundary face of $$\sigma$$ to make $$\partial\alpha$$ equal to $$0$$. Now the idea is that every other $$n$$-simplex in the same connected component as $$\sigma$$ can be reached by repeatedly taking $$n$$-simplices that share boundary faces in this way, and so every $$n$$-simplex in the same connected component as $$\sigma$$ must appear in $$\alpha$$. But since no connected component of $$U$$ is compact, this would mean infinitely many $$n$$-simplices must appear in $$\alpha$$, which is a contradiction. Thus no $$n$$-simplex $$\sigma$$ can have a nonzero coefficient in $$\alpha$$, so $$\alpha=0$$.

It remains to prove rigorously that every $$n$$-simplex in the same connected component as $$\sigma$$ can be reached from $$\sigma$$ by repeatedly taking $$n$$-simplices that share boundary faces. This follows from a connectedness argument using the fact that simplices of dimension less than $$n-1$$ can be "ignored" for the purposes of connectedness. Let $$V$$ be the connected component of $$U$$ containing $$\sigma$$, let $$V_0$$ be the $$(n-2)$$-skeleton of $$V$$, and let $$S$$ be the set of $$n$$-simplices and $$(n-1)$$-simplices of the triangulation which are contained in $$V$$. Define an equivalence relation $$\sim$$ on $$S$$ by saying two simplices are equivalent if one can be reached from the other by repeatedly passing between $$n$$-simplices and their boundary faces. If $$A\subseteq S$$ is an equivalence class with respect to $$\sim$$, let $$V(A)\subseteq V$$ be the union of the interiors of all the simplices of $$A$$. Note that $$V(A)$$ is connected and open in $$V\setminus V_0$$. Moreover, these sets $$V(A)$$ form a partition of $$V\setminus V_0$$ into open sets as $$A$$ ranges over all the equivalence classes in $$S$$. So to conclude that there is only one such equivalence class (so every $$n$$-simplex in $$V$$ can be reached from $$\sigma$$ by repeatedly passing through shared boundary faces), it suffices to show that $$V\setminus V_0$$ is connected.

To prove that $$V\setminus V_0$$ is connected, take two points $$p,q\in V\setminus V_0$$. Since $$V$$ is a connected open subset of $$\mathbb{R}^n$$, we can connect $$p$$ and $$q$$ by a piecewise linear path in $$V$$. Now the idea is that you can perturb the sequence of endpoints $$(p,x_1,\dots,x_m,q)$$ of the linear pieces of this path to avoid $$V_0$$. There are various ways to formulate this argument; for instance, you can consider the sequence $$(x_1,\dots,x_m)$$ of intermediate endpoints of the linear pieces as an element of $$(\mathbb{R}^n)^m$$. Then, for any linear simplex $$\sigma\subset\mathbb{R}^n$$ of dimension at most $$n-2$$, the set of such sequences in $$(\mathbb{R}^n)^m$$ for which the piecewise linear path will intersect $$\sigma$$ is closed and nowhere dense (this boils down to the fact that given two points in $$\mathbb{R}^n$$, you can perturb them so that the line segment between them does not pass through $$\mathbb{R}^{n-2}$$; by ignoring the first $$n-2$$ coordinates, you can reduce this to the case $$n=2$$ where it is obvious). Since $$V_0$$ is a union of countably many such simplices, by the Baire category theorem, the subset of $$(\mathbb{R}^n)^m$$ consisting of sequences that give piecewise linear paths that avoid $$V_0$$ is dense. So, we can pick such a sequence that is sufficiently close to $$(x_1,\dots,x_m)$$ so that the piecewise linear path remains in $$V$$, and thus we get a path between $$p$$ and $$q$$ in $$V\setminus V_0$$.